### Problem Statement (9.31 and 9.32)Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x-axis.### Diagram DescriptionThe diagram is labeled as Fig. P9.32. It features a composite shape on a coordinate plane with x and y axes. The structure consists of a large rectangle with two smaller rectangles subtracted from its interior, forming a C-shaped area. #### Dimensions:- The exterior rectangle's total width is 8 inches (composed of two 3-inch segments and two 2-inch segments).- The exterior rectangle's total height is 4 inches (two segments of 2 inches each).- The smaller rectangle on the left has dimensions of 6 inches by 2 inches.- The smaller rectangle on the right has dimensions of 4 inches by 2 inches.- The shaded area's top is positioned at y = 2 inches and its bottom is at y = -2 inches.- The right edge of both inner rectangles aligns with x = 2 inches.The point \(O\) marks the origin at the center of the structure, aligned with the x and y axes.### Task- Calculate the moment of inertia for the shaded area concerning the x-axis.- Calculate the radius of gyration for the shaded area concerning the x-axis.

The schematic diagram of the area is shown below:The moment of inertia of the shaded area with respect to the x-axis can be determined by subtracting the moment of inertia of the rectangle PQRS and the moment of inertia of the rectangle EFGH from the moment of inertia of the rectangle ABCD. Ix=IABCD,x-IPQRS,x-IEFGH,xIx=112AB×AD3-112PS×QP3-112FG×FE3 Here, the moment of inertia of the rectangle ABCD with respect to the x-axis is IABCD,x, the moment of inertia of the rectangle PQRS with respect to the x-axis is IPQRS,x and the moment of inertia of the rectangle EFGH with respect to the x-axis is IEFGH,x. Substitute all the values in the above expression, Ix=11212 in×8 in3-1125 in×4 in31122 in×6 in3Ix=512 in3-26.67 in3-36 in3Ix=449.33 in4Calculate the area of the shaded region. A=AreaABCD-AreaPQRS-AreaEFGHA=AB×AD-PS×QP-FG×FEA=12 in×8 in-5 in×4 in-2 in×6 inA=96 in2-20 in2-12 in2A=64 in2 Calculate the radius of gyration of the shaded area with respect to the x-axis as follows: k=IAk=449.33 in464 in2k≈2.65 in Therefore, the moment of inertia of the shaded area with respect to the x-axis is 449.33 in4 and the radius of gyration of the shaded area with respect to the x-axis is 2.65 in.

Engineering

Mechanical Engineering

9.31 and 9.32 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis. 2 in. 2 in. 3 in. 3 in.- 1 in. 3 in. 2 in. 2 in. 3 in. 1 in. 1 in. -1 in. Fig. P9.32.

9.31 and 9.32 Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x axis. 2 in. 2 in. 3 in. 3 in.- 1 in. 3 in. 2 in. 2 in. 3 in. 1 in. 1 in. -1 in. Fig. P9.32.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provide resistance in changing its state of velocity or motion for coming to a state of rest. In other words, Newton's first law of motion describes the law of inertia of a rigid body.

Question

### Problem Statement (9.31 and 9.32)

Determine the moment of inertia and the radius of gyration of the shaded area with respect to the x-axis.

### Diagram Description

The diagram is labeled as Fig. P9.32. It features a composite shape on a coordinate plane with x and y axes. The structure consists of a large rectangle with two smaller rectangles subtracted from its interior, forming a C-shaped area.

#### Dimensions:
- The exterior rectangle's total width is 8 inches (composed of two 3-inch segments and two 2-inch segments).
- The exterior rectangle's total height is 4 inches (two segments of 2 inches each).
- The smaller rectangle on the left has dimensions of 6 inches by 2 inches.
- The smaller rectangle on the right has dimensions of 4 inches by 2 inches.
- The shaded area's top is positioned at y = 2 inches and its bottom is at y = -2 inches.
- The right edge of both inner rectangles aligns with x = 2 inches.

The point \(O\) marks the origin at the center of the structure, aligned with the x and y axes.

### Task

- Calculate the moment of inertia for the shaded area concerning the x-axis.
- Calculate the radius of gyration for the shaded area concerning the x-axis.

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